- CFA Exams
- CFA Level I Exam
- Study Session 3. Quantitative Methods (2)
- Reading 9. Common Probability Distributions
- Subject 5. The Binomial Distribution
CFA Practice Question
The sponsors of a well-known charity came up with a unique idea to attract wealthy patrons to a $500-a-plate dinner. After the dinner, it was announced that each patron attending could buy a set of 20 tickets for the gaming tables. The chance of winning a prize for each of the 20 plays is 50-50. If you bought a set of 20 tickets, what is the chance that you will win 15 or more prizes?
A. 0.006
B. 0.250
C. 0.021
Explanation: This is a binomial probability. The probability of getting r successes out of n trials, where the probability of success each trial is p and probability of failure each trial is q (where q = 1-p), is given by: n!(pr)[q(n-r)]/r!(n-r)!. Here, n = 20, p = 0.5 and q = 0.5 and r = 15,16,17,18,19,20. Therefore, we have:
P(16) = 20!(0.516)(0.54)/16!4! = 0.0046
P(17) = 0.0011
P(18) = 0.0002
P(19) = 0.00002
P(20) = 0.000001
P(15) = 20!(0.515)(0.55)/15!5! = 0.0148
P(16) = 20!(0.516)(0.54)/16!4! = 0.0046
P(17) = 0.0011
P(18) = 0.0002
P(19) = 0.00002
P(20) = 0.000001
The sum is 0.0207.
User Contributed Comments 6
| User | Comment |
|---|---|
| baddabing | Quite easy isn't it! |
| Xocrevilo | The answer provided is too long-winded! You can deduce the answer after just calculating the first term: P(15)=... 0.0148. 0.0148 is greater than 0.006, thus reject answer A. We also know that the probabilities of P(16), P(17), P(18), P(19) and P(20) will all be smaller than 0.0148, but in sum are unlikely to add up to 0.75, thus reject answers B. |
| chamad | thank you for the tip. For sure the calculation is time consuming in the exam. |
| aakash1108 | .........the problem would begin when the other 3 options are greater than but close to 0.0148. |
| SCBAnalyst | How do you calculate this using a BAII Plus Calculator |
| at0899 | if thats the professional version... (20 nCr 15 )[(.5)^15][(.5)^5] => .0148 |